Advertisements
Advertisements
प्रश्न
Figure given below shows a velocity-time graph for a car starting from rest. The graph has three parts AB, BC and CD.
Compare the distance travelled in part BC with the distance travelled in part AB.
Advertisements
उत्तर
Distance travelled in part BC = Area of the rectangle tBC2t = base × height.
= (2t - t) × vo
= vot
Distance travelled in part AB = Area of the triangle ABt
= (1/2) × base × height
= (1/2) × t × vo
= (1/2) vo t
Therefore, distance travelled in part BC:distance travelled in part AB :: 2:1.
APPEARS IN
संबंधित प्रश्न
If the displacement of an object is proportional to the square of time, then the object is moving with :
When is the magnitude of displacement equal to the distance?
Figure shows the displacement-time graph of two vehicles A and B moving along a straight road. Which vehicle is moving faster? Give reason.
A car starting from rest accelerates uniformly to acquire a speed 20 km h-1 in 30 min. The distance travelled by a car in this time interval will be _____________
A body starts from rest with uniform acceleration 2 m s-2. Find the distance covered by the body in 2 s.
From the displacement – time graph shown given below calculate :
- Velocity between 0 – 2 s.
- Velocity between 8 s – 12 s.
- Average velocity between 5 s – 12 s.
A satellite makes a complete round along its circular orbit. What is its displacement?
The figure represents graphically the velocity of a car moving along a straight road over a period of 100 hours.
Calculate the distance travelled in the last 40 h.
Using the following data, draw time - displacement graph for a moving object:
| Time (s) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 |
| Displacement (m) | 0 | 2 | 4 | 4 | 4 | 6 | 4 | 2 | 0 |
Use this graph to find average velocity for the first 4 s, for the next 4 s and for the last 6 s.
A man leaves his house at 6.30 a.m. for a morning walk and returns back at 7.30 a.m. after covering 4 km. The Displacement covered by him is ______.
